Math / Geometry

Question $(x+3)^{2}+(y-4)^{2}=9$
A circle in the $x y$ -plane has the equation shown. If the $x$ -coordinate of a point on the circle is -3 , what is a possible corresponding $y$ -coordinate? $\square$

Studdy Solution

STEP 1

1. The equation $(x+3)^{2}+(y-4)^{2}=9$ represents a circle in the $xy$ -plane.
2. The center of the circle is at $(-3, 4)$ .
3. The radius of the circle is $3$ .
4. We need to find the $y$ -coordinate of a point on the circle where the $x$ -coordinate is $-3$ .

STEP 2

1. Substitute the given $x$ -coordinate into the circle's equation.
2. Simplify the equation to solve for $y$ .
3. Solve for the possible $y$ -coordinates.

STEP 3

Substitute the given $x$ -coordinate $-3$ into the circle's equation:
$((-3)+3)^{2}+(y-4)^{2}=9$

STEP 4

Simplify the equation:
$(0)^{2}+(y-4)^{2}=9$ $(y-4)^{2}=9$

STEP 5

Solve for $y$ by taking the square root of both sides:
$y-4 = \pm 3$
Solve for $y$ :
1. $y - 4 = 3$ $ y = 7 \]
2. $y - 4 = -3$ $ y = 1 \]
The possible $y$ -coordinates are $y = 7$ and $y = 1$ .
A possible corresponding $y$ -coordinate is:
$\boxed{7} \quad \text{or} \quad \boxed{1}$

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